Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int main() {
int n;
cin>>n;
double a[n];
for(int i=0;i<n;i++)
cin>>a[i];

double dp[n+1][n+1];
dp[0][0]=1;
for(int i=1;i<=n;i++)
dp[0][i]=0;
for(int i=1;i<=n;i++)
{
    for(int j=0;j<=i;j++)     {
    
    if(j)
    dp[i][j]=dp[i-1][j-1]*a[i-1];
    if(i!=j)
    dp[i][j]+=dp[i-1][j]*(1-a[i-1]);
    
}
    
}
double sum=0;
for(int i=n/2+1;i<=n;i++)
sum+=dp[n][i];
printf("%.10f",sum);



return 0;
}